Q.Radiation of wavelength 4500 Å is incident on a metal having work function 2.0 eV. Due to the presence of a magnetic field B, the most energetic photoelectrons emitted in a direction perpendicular to the field move along a circular path of radius 20 cm. What is the value of the magnetic field B? [Ans. : $1.473\times10^{-5}$ T]

[!TLDR] $KE_{max}=hc/\lambda-\phi_0\approx0.7625$ eV $=1.220\times10^{-19}$ J, giving speed $v=\sqrt{2KE_{max}/m_e}\approx5.18\times10^{5}$ m/s; then from circular motion in a magnetic field, $B=\df …

First find the maximum kinetic energy of the photoelectrons using Einstein's photoelectric equation. With $\lambda=4500$ Å and $hc/e=12431$ eV Å:\n\n$\frac{hc}{\lambda} = \frac{12431}{4500} \approx 2.7625\text{ eV}$\n\n$KE_{max} = \frac{hc}{\lambda} - \phi_0 = 2.7625 - 2.0 = 0.7625\text{ eV} = 0.7625\times1.6\times10^{-19}\text{ J} \approx 1.220\times10^{-19}\text{ J}$\n\nThe speed of the most energetic photoelectrons follows from $KE_{max}=\frac{1}{2}m_ev^2$:\n\n$v = \sqrt{\frac{2\,KE_{max}}{m_e}} = \sqrt{\frac{2\times1.220\times10^{-19}}{9.11\times10^{-31}}} = \sqrt{2.678\times10^{11}} \approx 5.175\times10^5\text{ m/s}$\n\nWhen this electron moves PERPENDICULAR to the magnetic field B, it experiences a magnetic force $evB$ that acts as the centripetal force keeping it on a circular path of radius r, so $ev …

Find $KE_{max}$ from Einstein's photoelectric equation, convert to speed via $KE_{max}=\frac{1}{2}m_ev^2$, then use the circu …

Forgetting to convert $KE_{max}$ from eV to joules before using it in $v=\sqrt{2KE_{max}/m_e}$ (where $m_e$ is in SI …